Question

Ratio of $C_P$ and $C_V$ of gas '$X$' is 1.4. The number of atoms of the gas '$X$' present in $11.2$ litres of at NTP will be:

• A. $6.02\times {10}^{23}$
• B. $5.61\times {10}^{23}$
• C. $3.01\times {10}^{23}$
• D. $2.01\times {10}^{23}$

Answer 1

Answer: (A)

$6.02\times {10}^{23}$

$\frac{C_p}{C_v}=\gamma =1.4$

Hence, the gas is diatomic.

Number of atoms $=$ atoms in one molecule of gas $\times$ mole$\times N_A$

$=2\times \frac{V}{22.4}\times N_A$

$=2\times \frac{11.2}{22.4}\times 6.02\times {10}^{23}=6.02\times {10}^{23}$